in this video we're analyzing a perfectly elastic collision between two carts and they bounce off
of each other using a hooke's law spring and the real emphasis in this problem is to use the center
of mass reference frame to analyze the collision so in part a we're looking at the center of mass
velocity and then transforming the initial velocities of the carts into the center of
mass frame in part b we take advantage of the center of mass frame to figure out the maximum
compression of the spring during the collision and in part c we take advantage of the center of
mass frame to find the final velocities of each cart within the center of mass frame and then
transform those into the lab frame to kind of get the final answer for the outcome of the collision
so this idea of center of mass velocity this comes from a fact about the total momentum
of a system of moving objects and i'll post a link to the video where this was derived
the total momentum i'll call that p-net is equal to the total mass of the system so the mass of
all the objects in the system added together and we usually use a capital m for that multiplied by
the center of mass velocity so as these two carts move along the track the center of mass changes
position and the center of mass velocity is just tracking how fast the center of mass moves so the
way we use this is by plugging in the masses and velocities of the initially moving carts to get
the net momentum then we're able to solve for the center of mass velocity so the cart on the left
that's 250 grams or 0.25 kilograms moving with a positive 5.1 meters per second the car on the
right we're going to give that a minus sign for the velocity so we'll just move that out in front
400 grams is 0.4 kilograms and again the speed is 1.5 meters per second the minus sign takes
care of the direction on the right hand side of this equation we have the total mass which is
650 grams or 0.65 kilograms and the only unknown here is the center of mass velocity and when we
solve for the center of mass velocity we get 1.038 just keeping a little extra precision today 1.038
meters per second so that's tracking to the right at 1.038 meters per second so the next part of
the question is a little bit tricky we have to find the initial velocity of each of these carts
within the center of mass reference frame what that means is we're going to run along to the
right as this collision is occurring we're going to run at 1.038 meters per second to the right
this makes the center of mass of the system look stationary in the new reference frame
so just to show our perspective here we are now running to the right at 1.038 meters per second
and observing the initial and final states of this collision
so from this new perspective of running to the right at 1.038 meters per second the cart on
the left the 250 gram cart that's going to look like it's moving slower it was moving 5.1 meters
per second to the right in the lab frame now we're sort of attempting to catch up to it we're
running to the right and that makes it look 1.038 meters per second slower as it moves to the right
so this is going to be our new initial velocity i'll call this v1 initial it's going to be 5.1
minus 1.038 this gives us 4.062 meters per second now with the 400 gram cart we get the
opposite effect so we're running toward that it was moving to the left and we're running to the
right that makes it look like it's going faster to the left so if i just speak in terms of magnitudes
here in other words i'm writing down the speed of this cart and not worrying about the sign for the
direction right now v2 initial is going to be this initial speed of 1.5 meters per second to the left
plus an additional 1.038 meters per second added to that magnitude and this gives us 2.538
meters per second so in parts b and c we're going to get some useful results out of applying the
center of mass frame to the problem and the real power of the center of mass frame is that the net
momentum is zero within that frame and we can see that from our original equation for the center of
mass velocity the net momentum is equal to the total mass times the center of mass velocity
in the center of mass frame we're observing the experiment from a frame that tracks
with the center of mass and that means the center of mass velocity is zero in that frame
and that means the net momentum is zero so i want to take a moment just to show that just
to verify that the net momentum is zero within the center of mass frame and then we're going to
apply that fact to quickly solve the rest of the problem so within the center of mass reference
frame the momentum of the first cart and i mean the 250 gram one is going to be 0.25 kilograms
and we're going to take our center of mass initial velocity 4.062 meters per second and this gives us
1.016 kilogram meters per second for the rightward momentum of the first cart within the center of
mass reference frame we do the same thing for the second cart p2 initial that's 0.4 kilograms
and it has a leftward momentum and again we're just talking about the magnitude of that right now
it had a speed of 2.538 meters per second and this gives us 1.015 kilogram meters per second
and of course if we were including the direction that would be negative so these two quantities are
practically equal there's a little discrepancy due to rounding differences i have a positive momentum
to the right for that first cart that has the same magnitude of the leftward pointing momentum
for the second cart and my net momentum turns out to be zero and the reason this is exciting
is that collisions where the net momentum is zero turn out to be particularly easy to analyze
so again in part b we're trying to find the maximum compression of this
spring and when does this occur it occurs when the kinetic energy is at a minimum and in fact
because i know the net momentum is zero within this reference frame i can figure out the minimum
kinetic energy it's when both carts are completely stationary so it's really what we're looking at
in the second picture on the left during this impact i actually have v1 is equal to zero within
the center of mass reference frame v2 equals zero within the center of mass reference frame
i know this is going to happen because i see immediately that that's one way of having zero
momentum within the center of mass reference frame so within this frame the kinetic energy is zero
at the moment of maximum compression of the spring and all of that kinetic energy has gone into the
potential energy within the spring so we have an energy conservation problem where we start with
our initial kinetic energies within the center of mass reference frame using the speeds that we
found in part a we set that equal to the spring potential energy with an unknown compression
length and we solve for the compression length so symbolically real quick we have the potential
energy in the spring one half kx squared is going to be equal to our initial kinetic energy in the
first mass plus the initial kinetic energy in the second mass within the center of mass reference
frame so we plug the numbers in we were told the spring constant was 70 newtons per meter
the compression distance x is the only unknown here the initial kinetic energy of the first
mass is going to be one half times its mass that was 0.25 times its speed squared 4.062 squared
and then i have a one-half times 0.4 0.4 for the second mass times 2.538
for the speed all squared so of course we can multiply through by 2 to clean things up a little
bit we smash all the numbers together on the right divide by 70 and square root the result and we'll
get our compression distance and we end up with a compression distance of 0.309 meters and it might
be nicer to write that in centimeters that's 30.9 centimeters finally in part c we're going to apply
the power of the center of mass reference frame to basically guess the final velocities of these
carts within the center of mass frame and then we're going to transform back to the lab frame
so again the net momentum of the system is zero within the center of mass reference frame and
what we're trying to do is find the outcome of a perfectly elastic collision within that frame
we know the initial momentum is zero we know the kinetic energy is conserved and if we can guess
a solution that makes it so the final momentum is equal to the initial momentum and the final
kinetic energy is equal to the initial kinetic energy that solution is guaranteed to be correct
and there is an easily guessable solution all we have to do is send the initially rightward
moving mass back to the left with the same exact speed it started with so 4.062 meters per second
within the center of mass reference frame send the heavier mass the 400 gram one back to the right
with exactly the same speed that it started with in the center of mass reference frame that simply
changes the sign on each of their momenta making it so the net momentum is still zero
and by giving the same speed to each of the same masses that makes the kinetic energy come out
the same so that's really the whole point of the center of mass reference frame is it makes elastic
collisions guessable so within the center of mass reference frame and i'm going to go ahead and talk
about these including their direction now so i'll go ahead and put a vector hat on v1 final that's
going to be negative 4.062 meters per second just meaning 4.062 meters per second to the left
v2 final it was initially moving to the left before the collision occurs it's now going to
move to the right so i'm going to write plus 2.538 meters per second again we've already verified
that the momentum of these two things turns out to be the same so all we've done here is reverse
the directions we have the same magnitudes of the momentum opposite signs so the momentum still adds
to zero secondly these objects have the same mass as they originally did and the same speed as they
originally did so the kinetic energy is also the same so there's our guessable solution to
the elastic collision within the center of mass reference frame now the challenge is to convert
back to the lab frame and remember the center of mass reference frame is the way we see the world
when we're running to the right at 1.038 meters per second v1 final has a leftward velocity of
4.062 meters per second from that perspective and remember a leftward velocity is exaggerated when
you're running to the right so we need to reduce that effect we're going to reduce the magnitude of
the speed by 1.038 meters per second algebraically what we're doing is adding the 1.038 to this
velocity and we end up with a velocity in the lab frame of negative 3.02 just keeping three sig figs
for our final answer and of course the units are meters per second there now what about v2 final so
within the center of mass frame that's when we're running to the right at 1.038 meters per second
we see v2 moving to the right at 2.538 meters per second now remember if we're running to the right
a rightward velocity has been underestimated so we need to add in that magnitude of 1.038
meters per second so we have 2.538 plus a 1.038 to figure out what it looks like from the lab frame
and again to three significant digits i have 3.58 meters per second now if this all seems a
little bit suspicious to you changing into a new reference frame and turning an elastic collision
into a guessable problem and then changing back remember that any solution that conserves momentum
and conserves kinetic energy is going to be the right answer for this elastic collision
so the way to check your work on this problem is to look at the final velocities for the lab
reference frame so again we end up with the smaller cart reflecting and moving to the left
at 3.02 meters per second the larger cart also reflects from its original direction
and it ends up moving at 3.58 meters per second to the right so to check our answer here we just
compute the total initial momentum within the lab frame compute the final momentum within the lab
frame and make sure they come out the same and then compute the total initial kinetic energy
in the lab frame the final kinetic energy in the lab frame and make sure those come out the same
so i actually ran those numbers this afternoon and it does all check out just one final comment
here i think this problem is such a great window into the way that physicists think about the world
if we just did the lab frame analysis of this problem to figure out the final velocities of
our masses there would have been a heck of a lot of algebra in there there's a non-linear system
of equations that has to be solved so instead we opted for a trade-off that was a little bit
abstract and difficult we had to imagine what the world looked like running to the right at
1.038 meters per second but the trade-off is that all the algebra went away and the answer
to the problem became guessable then we just had to transform back into the lab frame and we got
our final answers so i hope this video makes it a little more likely for you to consider
using a transformation to a new reference frame to analyze a collision problem i posted a link to a
playlist on the left with some more center of mass and collision problems and i'll see you next time